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Calorimetry - determining Determining enthalpy and internal energy of formation for K2Opotassium superoxide

Given $2.14\ \mathrm g$ of $\ce{K(s)}$, determine $\Delta H^\circ_\mathrm{f,m}$ and $\Delta U^\circ_\mathrm{f,m}$ for $\ce{K2O}$.

We know:

  • The calorimeter's constant: $1849\ \mathrm{J\cdot K^{-1}}$
  • The mass of water inside it: $1450\ \mathrm g$
  • The change in temperature: $2.62\ \mathrm K$
  • The end product is $\ce{K2O}$

We know:

  • The calorimeter's constant: $1849\ \mathrm{J\cdot K^{-1}}$
  • The mass of water inside it: $1450\ \mathrm g$
  • The change in temperature: $2.62\ \mathrm K$
  • The end product is $\ce{K2O}$

The process should be: determining mol of $\ce{K(s)}$, which is $0.054\ \mathrm{mol}$. Then we obtain the amount of energy absorbed by the calorimeter/water. Here are the issues. I don't know which one changed its temperature by $2.62\ \mathrm K$ or if the calorimeter's constant already considers the water. Either way, the amount of energy released by each mol of potassium is extravagant; according to the result it is $322\ \mathrm{kJ}$ or $96.8\ \mathrm{kJ}$.

Continuing with this reasoning, the $\Delta U^\circ_\mathrm{f,m}$ should be $755\ \mathrm{kJ}$ or $193\ \mathrm{kJ}$, right? What do we need to obtain $\Delta H^\circ_\mathrm{f,m}$?

Calorimetry - determining enthalpy and internal energy of formation for K2O

Given $2.14\ \mathrm g$ of $\ce{K(s)}$, determine $\Delta H^\circ_\mathrm{f,m}$ and $\Delta U^\circ_\mathrm{f,m}$ for $\ce{K2O}$.

We know:

  • The calorimeter's constant: $1849\ \mathrm{J\cdot K^{-1}}$
  • The mass of water inside it: $1450\ \mathrm g$
  • The change in temperature: $2.62\ \mathrm K$
  • The end product is $\ce{K2O}$

The process should be: determining mol of $\ce{K(s)}$, which is $0.054\ \mathrm{mol}$. Then we obtain the amount of energy absorbed by the calorimeter/water. Here are the issues. I don't know which one changed its temperature by $2.62\ \mathrm K$ or if the calorimeter's constant already considers the water. Either way, the amount of energy released by each mol of potassium is extravagant; according to the result it is $322\ \mathrm{kJ}$ or $96.8\ \mathrm{kJ}$.

Continuing with this reasoning, the $\Delta U^\circ_\mathrm{f,m}$ should be $755\ \mathrm{kJ}$ or $193\ \mathrm{kJ}$, right? What do we need to obtain $\Delta H^\circ_\mathrm{f,m}$?

Determining enthalpy and internal energy of formation for potassium superoxide

Given $2.14\ \mathrm g$ of $\ce{K(s)}$, determine $\Delta H^\circ_\mathrm{f,m}$ and $\Delta U^\circ_\mathrm{f,m}$ for $\ce{K2O}$.

We know:

  • The calorimeter's constant: $1849\ \mathrm{J\cdot K^{-1}}$
  • The mass of water inside it: $1450\ \mathrm g$
  • The change in temperature: $2.62\ \mathrm K$
  • The end product is $\ce{K2O}$

The process should be: determining mol of $\ce{K(s)}$, which is $0.054\ \mathrm{mol}$. Then we obtain the amount of energy absorbed by the calorimeter/water. Here are the issues. I don't know which one changed its temperature by $2.62\ \mathrm K$ or if the calorimeter's constant already considers the water. Either way, the amount of energy released by each mol of potassium is extravagant; according to the result it is $322\ \mathrm{kJ}$ or $96.8\ \mathrm{kJ}$.

Continuing with this reasoning, the $\Delta U^\circ_\mathrm{f,m}$ should be $755\ \mathrm{kJ}$ or $193\ \mathrm{kJ}$, right? What do we need to obtain $\Delta H^\circ_\mathrm{f,m}$?

3 typography corrected
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Given $2.14g$ of $K_{(s)}$, determine $\Delta Hº_{f,m}$ and $\Delta Uº_{f,m}$ for $K_2O$.

Given $2.14\ \mathrm g$ of $\ce{K(s)}$, determine $\Delta H^\circ_\mathrm{f,m}$ and $\Delta U^\circ_\mathrm{f,m}$ for $\ce{K2O}$.

We know:

  • The calorimeter's constant: $1849J•K^{-1}$$1849\ \mathrm{J\cdot K^{-1}}$
  • The mass of water inside it: $1450g$$1450\ \mathrm g$
  • The change in temperature: $2.62K$$2.62\ \mathrm K$
  • The end product is $K_2O$$\ce{K2O}$

The process should be: determining mol of $K_{(s)}$$\ce{K(s)}$, which is $0.054mol$$0.054\ \mathrm{mol}$. Then we obtain the amount of energy absorbed by the calorimeter/water. Here are the issues. I don't know which one changed its temperature by $2.62K$$2.62\ \mathrm K$ or if the calorimeter's constant already considers the water. Either way, the amount of energy released by each mol of potassium is extravagant; according to the result it is $322kJ$$322\ \mathrm{kJ}$ or $96.8kJ$$96.8\ \mathrm{kJ}$.

Continuing with this reasoning, the $\Delta Uº_{f,m}$$\Delta U^\circ_\mathrm{f,m}$ should be $755kJ$$755\ \mathrm{kJ}$ or $193kJ$$193\ \mathrm{kJ}$, right? What do we need to obtain $\Delta Hº_{f,m}$$\Delta H^\circ_\mathrm{f,m}$?

Given $2.14g$ of $K_{(s)}$, determine $\Delta Hº_{f,m}$ and $\Delta Uº_{f,m}$ for $K_2O$.

We know:

  • The calorimeter's constant: $1849J•K^{-1}$
  • The mass of water inside it: $1450g$
  • The change in temperature: $2.62K$
  • The end product is $K_2O$

The process should be: determining mol of $K_{(s)}$, which is $0.054mol$. Then we obtain the amount of energy absorbed by the calorimeter/water. Here are the issues. I don't know which one changed its temperature by $2.62K$ or if the calorimeter's constant already considers the water. Either way, the amount of energy released by each mol of potassium is extravagant; according to the result it is $322kJ$ or $96.8kJ$.

Continuing with this reasoning, the $\Delta Uº_{f,m}$ should be $755kJ$ or $193kJ$, right? What do we need to obtain $\Delta Hº_{f,m}$?

Given $2.14\ \mathrm g$ of $\ce{K(s)}$, determine $\Delta H^\circ_\mathrm{f,m}$ and $\Delta U^\circ_\mathrm{f,m}$ for $\ce{K2O}$.

We know:

  • The calorimeter's constant: $1849\ \mathrm{J\cdot K^{-1}}$
  • The mass of water inside it: $1450\ \mathrm g$
  • The change in temperature: $2.62\ \mathrm K$
  • The end product is $\ce{K2O}$

The process should be: determining mol of $\ce{K(s)}$, which is $0.054\ \mathrm{mol}$. Then we obtain the amount of energy absorbed by the calorimeter/water. Here are the issues. I don't know which one changed its temperature by $2.62\ \mathrm K$ or if the calorimeter's constant already considers the water. Either way, the amount of energy released by each mol of potassium is extravagant; according to the result it is $322\ \mathrm{kJ}$ or $96.8\ \mathrm{kJ}$.

Continuing with this reasoning, the $\Delta U^\circ_\mathrm{f,m}$ should be $755\ \mathrm{kJ}$ or $193\ \mathrm{kJ}$, right? What do we need to obtain $\Delta H^\circ_\mathrm{f,m}$?

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Problem with a calorimeter Calorimetry (theoretical)- determining enthalpy and internal energy of formation for K2O

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